******************************************************************************* Examples of STANJAN runs: This example shows how STANJAN can be used to calculate the adiabatic flame temperature in a gas turbine engine combustor, and then the composition after isentropic expansion in the turbine. Note: Your output may differ slightly due to version upgrade. ------------------------------------------------------------------------------- Step1: Get the enthalpy of the reactants by a run at the combustor inlet state. Here we took T = 400 K, P = 6 atm, and CH4 + 2*O2 + 7.52*N2 The data are taken from species data file COMB.SUD. The STANJAN output follows. Independent relative element atom population potential H 4.00000000E+00 -11.4094 O 4.00000000E+00 -12.3340 N 1.50400000E+01 -10.8571 Dependent atom C 1.00000000E+00 Composition at T = 400.00 K P = 6.000E+00 atmospheres species mol fraction mol fraction mass fraction mols* in the phase in mixture in mixture Phase 1: Molal mass, g/mol = 27.633 CH4 .95057E-01 .95057E-01 .55187E-01 1.00000E+00 O2 .19011E+00 .19011E+00 .22015E+00 2.00000E+00 N2 .71483E+00 .71483E+00 .72466E+00 7.52000E+00 * Species mols for the atom populations in mols. Mixture properties: molal mass = 27.633 kg/kmol T = 400.000 K P = 6.0795E+05 Pa V = 1.9796E-01 m**3/kg U =-2.6694E+05 J/kg H =-1.4659E+05 J/kg S = 7.0201E+03 J/kg-K Made 0 (T,P) iterations; 1 EQUIL iterations. ------------------------------------------------------------------------------- Step 2: Get the adiabatic flame temperature by finding the state of the products at the same enthalpy (H) and pressure (P) as the last run. The species in COMB.SUD are allowed as products. The STANJAN output follows. Independent relative element atom population potential C 1.00000000E+00 -20.4676 H 4.00000000E+00 -11.9990 O 4.00000000E+00 -16.5667 N 1.50400000E+01 -12.9931 Composition at T = 2316.31 K P = 6.000E+00 atmospheres species mol fraction mol fraction mass fraction mols* in the phase in mixture in mixture Phase 1: Molal mass, g/mol = 27.466 C .55358E-16 .55358E-16 .24209E-16 5.85920E-16 CH4 .25808E-15 .25808E-15 .15074E-15 2.73153E-15 CO .75620E-02 .75620E-02 .77120E-02 8.00375E-02 CO2 .86918E-01 .86918E-01 .13927E+00 9.19962E-01 H .22987E-03 .22987E-03 .84361E-05 2.43302E-03 H2 .28857E-02 .28857E-02 .21181E-03 3.05426E-02 H2O .18475E+00 .18475E+00 .12118E+00 1.95542E+00 HO .24233E-02 .24233E-02 .15006E-02 2.56492E-02 N .15822E-07 .15822E-07 .80693E-08 1.67459E-07 N2 .70944E+00 .70944E+00 .72358E+00 7.50883E+00 NO .21100E-02 .21100E-02 .23053E-02 2.23330E-02 NO2 .67724E-06 .67724E-06 .11345E-05 7.16807E-06 O .13180E-03 .13180E-03 .76780E-04 1.39501E-03 O2 .35539E-02 .35539E-02 .41404E-02 3.76148E-02 Phase 2: Molal mass, g/mol = .000 C(S) .00000E+00 .00000E+00 .00000E+00 .00000E+00 Phase 3: Molal mass, g/mol = .000 H2O(L) .00000E+00 .00000E+00 .00000E+00 .00000E+00 * Species mols for the atom populations in mols. Mixture properties: molal mass = 27.466 kg/kmol T = 2316.309 K P = 6.0795E+05 Pa V = 1.1534E+00 m**3/kg U =-8.4777E+05 J/kg H =-1.4659E+05 J/kg S = 9.3791E+03 J/kg-K Made 4 (T,P) iterations; 26 EQUIL iterations. ------------------------------------------------------------------------------ Step 3: Get the turbine exhaust temperature (isentropic process) by finding the equilibrium state of the same species at the same entropy (S) and a specified pressure (here P = 1 atm.). The STANJAN output, this time including the sound speed, follows. Independent relative element atom population potential C 1.00000000E+00 -24.7380 H 4.00000000E+00 -13.7096 O 4.00000000E+00 -18.3106 N 1.50400000E+01 -13.3409 Composition at T = 1674.93 K P = 1.000E+00 atmospheres species mol fraction mol fraction mass fraction mols* in the phase in mixture in mixture Phase 1: Molal mass, g/mol = 27.625 C .15608E-23 .15608E-23 .67862E-24 1.64248E-23 CH4 .93925E-20 .93925E-20 .54545E-20 9.88372E-20 CO .33225E-03 .33225E-03 .33688E-03 3.49629E-03 CO2 .94697E-01 .94697E-01 .15086E+00 9.96504E-01 H .16438E-05 .16438E-05 .59977E-07 1.72976E-05 H2 .20251E-03 .20251E-03 .14778E-04 2.13099E-03 H2O .18982E+00 .18982E+00 .12379E+00 1.99747E+00 HO .74204E-04 .74204E-04 .45683E-04 7.80848E-04 N .27866E-11 .27866E-11 .14130E-11 2.93239E-11 N2 .71458E+00 .71458E+00 .72462E+00 7.51956E+00 NO .84518E-04 .84518E-04 .91807E-04 8.89380E-04 NO2 .83678E-08 .83678E-08 .13936E-07 8.80548E-08 O .48685E-06 .48685E-06 .28197E-06 5.12314E-06 O2 .20673E-03 .20673E-03 .23946E-03 2.17541E-03 Phase 2: Molal mass, g/mol = .000 C(S) .00000E+00 .00000E+00 .00000E+00 .00000E+00 Phase 3: Molal mass, g/mol = .000 H2O(L) .00000E+00 .00000E+00 .00000E+00 .00000E+00 * Species mols for the atom populations in mols. Mixture properties: molal mass = 27.625 kg/kmol T = 1674.929 K P = 1.0133E+05 Pa V = 4.9750E+00 m**3/kg U =-1.7272E+06 J/kg H =-1.2231E+06 J/kg S = 9.3791E+03 J/kg-K Made 3 (T,P) iterations; 24 EQUIL iterations. Sound speed (isentropic) = 794.9 m/s +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ This example shows how to use STANJAN to calculate the state behind a Chapman- Jouguet detonation wave (sonic with respect to burned gas). ------------------------------------------------------------------------------ Step1: Establish the state of the unburned gas, here CH4 + 2(O2 + 3.76 N2) Independent relative element atom population potential H 4.00000000E+00 -13.6899 O 4.00000000E+00 -13.1601 N 1.50400000E+01 -11.6842 Dependent atom C 1.00000000E+00 Composition at T = 300.00 K P = 1.000E+00 atmospheres species mol fraction mol fraction mass fraction mols* in the phase in mixture in mixture Phase 1: molal mass = 27.633 kg/kmol CH4 .95057E-01 .95057E-01 .55187E-01 1.00000E+00 O2 .19011E+00 .19011E+00 .22015E+00 2.00000E+00 N2 .71483E+00 .71483E+00 .72466E+00 7.52000E+00 * Species mols for the atom populations in mols. Mixture properties: molal mass = 27.633 kg/kmol T = 300.00 K P = 1.0133E+05 Pa V = 8.9084E-01 m**3/kg U =-3.4581E+05 J/kg H =-2.5555E+05 J/kg S = 7.2459E+03 J/kg-K Made 0 (T,P) iterations; 1 equilibrium iterations; v 3.50 IBM-PC ------------------------------------------------------------------------------ Step 2: Set up the allowable species in the products, and run detonation: Independent relative element atom population potential C 1.00000000E+00 -18.6671 H 4.00000000E+00 -11.2399 O 4.00000000E+00 -15.8856 N 1.50400000E+01 -12.8072 Composition at T = 2781.32 K P = 1.709E+01 atmospheres species mol fraction mol fraction mass fraction mols* in the phase in mixture in mixture Phase 1: molal mass = 27.071 kg/kmol C .87604E-13 .87604E-13 .38868E-13 9.40720E-13 CH4 .27362E-13 .27362E-13 .16215E-13 2.93821E-13 CO .23589E-01 .23589E-01 .24407E-01 2.53303E-01 CO2 .69535E-01 .69535E-01 .11304E+00 7.46697E-01 H .17133E-02 .17133E-02 .63794E-04 1.83985E-02 HO .95081E-02 .95081E-02 .59734E-02 1.02101E-01 H2 .85651E-02 .85651E-02 .63784E-03 9.19753E-02 H2O .17207E+00 .17207E+00 .11451E+00 1.84777E+00 N .60727E-06 .60727E-06 .31423E-06 6.52108E-06 NO2 .41395E-05 .41395E-05 .70350E-05 4.44511E-05 NO .75774E-02 .75774E-02 .83994E-02 8.13693E-02 N2 .69650E+00 .69650E+00 .72074E+00 7.47929E+00 O .11952E-02 .11952E-02 .70637E-03 1.28340E-02 O2 .97377E-02 .97377E-02 .11510E-01 1.04567E-01 * Species mols for the atom populations in mols. Mixture properties: molal mass = 27.071 kg/kmol T = 2781.32 K P = 1.7313E+06 Pa V = 4.9340E-01 m**3/kg U = 1.8356E+04 J/kg H = 8.7257E+05 J/kg S = 9.4599E+03 J/kg-K Made 70 (T,P) iterations; 256 equilibrium iterations; v 3.50 IBM-PC Sound speed (isentropic) = 999.2 m/s *******************************************************************************